> >It seems clear that only the Condorcet winner (if one exists) will produce > >a Nash equilibrium. > >Are you suggesting that if there is no Condorcet winner there is no Nash >equilibrium, or that if there is a Condorcet winner all of the Nash >equilibria will elect the Condorcet winner, or both?
The latter and not the former. You are correct that Nash proved the former to be false, and in fact I provided an example of a Nash equilibrium where there was no Condorcet winner. I know of know proof that all Nash equilibria produce the Condorcet winner if one exists, but it seems intuitively clear if you play with examples (at least in sequential approval voting). Of course, if you can come up with a counter example, then it's not true. >I was wondering what the general conditions are (making certain >assumptions about voter behavior in the presence of repeated polling) for >equilibria where additional polls do not change the outcome. > >Those equilibria may or may not also be Nash equilibria, since the polling >assumptions that I've seen usually assume that people distinguish between >the top 2 in revising their strategies. That does not encompass all >possible strategic adjustments, however, so it is not entirely clear to me >that repeated polling will always find the Nash equilibrium and/or >Condorcet winner. I'd like to hope so, however. Anyway, it's an >interesting question to explore. Whether a result is a Nash equilibrium has nothing to do with how voters adjust their strategy from poll to poll. The only factor is the sincere preferences (payoffs) and the nature of the voting system. Of course, the nature of the voter strategy determines whether the polls will find or settle on a Nash equilibrium. For example, the Brams and Fishburn strategy assumptions cause the voters to leave the Nash equilibrium once they find it in my 10 AB>C, 8 BC>A, 6 C>AB, example. My instinct is that a Nash equilibrium associated with a Condorcet winner will tend to be more stable. But of course, it still depends on voter behavior. -Adam
